Dada una array arr[] que consiste en N enteros y un entero K , la tarea es encontrar la suma de los elementos de la array posibles recorriendo la array y sumando arr[i] / K , K número de veces al final de la array , si arr[i] es divisible por K . De lo contrario, detenga el recorrido.
Ejemplos:
Entrada: arr[] = {4, 6, 8, 2}, K = 2
Salida: 44
Explicación:
Se realizan las siguientes operaciones:
- Para arr[0](= 4): arr[0](= 4) es divisible por 2, por lo tanto agregue 4/2 = 2, 2 números de veces al final de la array. Ahora, la array se modifica a {4, 6, 8, 2, 2, 2}.
- Para arr[1](= 6): arr[1](= 6) es divisible por 2, por lo tanto agregue 6/2 = 3, 2 veces al final de la array. Ahora, la array se modifica a {4, 6, 8, 2, 2, 2, 3, 3}.
- Para arr[2](= 8): arr[2](= 8) es divisible por 2, por lo tanto agregue 8/2 = 4, 2 veces al final de la array. Ahora, la array se modifica a {4, 6, 8, 2, 2, 2, 3, 3, 4, 4}.
- Para arr[3](= 2): arr[3](= 2) es divisible por 2, por lo tanto agregue 2/2 = 1, 2 veces al final de la array. Ahora, la array se modifica a {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1}.
- Para arr[4](= 2): arr[4](= 2) es divisible por 2, por lo tanto agregue 2/2 = 1, 2 veces al final de la array. Ahora, la array se modifica a {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1}.
- Para arr[5](= 2): arr[5](= 2) es divisible por 2, por lo tanto agregue 2/2 = 1, 2 veces al final de la array. Ahora, la array se modifica a {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1}.
Después de completar los pasos anteriores, la suma de los elementos de la array es = 4 + 6 + 8 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 1 + 1 + 1 + 1 + 1 + 1 = 44.
Entrada: arr[] = {4, 6, 8, 9}, K = 2
Salida: 45
Enfoque ingenuo: el enfoque más simple para resolver el problema dado es atravesar la array dada y agregar el valor (arr[i]/K) K varias veces al final de la array. Después de completar los pasos anteriores, imprima la suma de los elementos de la array .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
vector<long long> v;
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
v.push_back(arr[i]);
}
// Traverse the vector
for (int i = 0;
i < v.size(); i++) {
// If v[i] is divisible by K
if (v[i] % K == 0) {
long long x = v[i] / K;
// Iterate over the range
// [0, K]
for (int j = 0; j < K; j++) {
// Update v
v.push_back(x);
}
}
// Otherwise
else
break;
}
// Traverse the vector v
for (int i = 0; i < v.size(); i++)
sum = sum + v[i];
// Return the sum of the updated array
return sum;
}
// Driver Code
int main()
{
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout << sum(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
ArrayList<Integer> v = new ArrayList<>();
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
v.add(arr[i]);
}
// Traverse the vector
for(int i = 0; i < v.size(); i++)
{
// If v[i] is divisible by K
if (v.get(i) % K == 0)
{
int x = v.get(i) / K;
// Iterate over the range
// [0, K]
for(int j = 0; j < K; j++)
{
// Update v
v.add(x);
}
}
// Otherwise
else
break;
}
// Traverse the vector v
for(int i = 0; i < v.size(); i++)
sum = sum + v.get(i);
// Return the sum of the updated array
return sum;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = arr.length;
System.out.println(sum(arr, N, K));
}
}
// This code is contributed by Kingash
Python3
# Python3 program for the above approach # Function to calculate sum of array # elements after adding arr[i] / K # to the end of the array if arr[i] # is divisible by K def summ(arr, N, K): # Stores the sum of the array sum = 4 v = [i for i in arr] # Traverse the vector for i in range(len(v)): # If v[i] is divisible by K if (v[i] % K == 0): x = v[i] // K # Iterate over the range # [0, K] for j in range(K): # Update v v.append(x) # Otherwise else: break # Traverse the vector v for i in range(len(v)): sum = sum + v[i] # Return the sum of the updated array return sum # Driver Code if __name__ == '__main__': arr = [ 4, 6, 8, 2 ] K = 2 N = len(arr) print(summ(arr, N, K)) # This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int[] arr, int N, int K)
{
// Stores the sum of the array
int sum = 0;
List<int> v = new List<int>();
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
v.Add(arr[i]);
}
// Traverse the vector
for (int i = 0; i < v.Count; i++) {
// If v[i] is divisible by K
if (v[i] % K == 0) {
int x = v[i] / K;
// Iterate over the range
// [0, K]
for (int j = 0; j < K; j++) {
// Update v
v.Add(x);
}
}
// Otherwise
else
break;
}
// Traverse the vector v
for (int i = 0; i < v.Count; i++)
sum = sum + v[i];
// Return the sum of the updated array
return sum;
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 4, 6, 8, 2 };
int K = 2;
int N = arr.Length;
Console.WriteLine(sum(arr, N, K));
}
}
// This code is contributed by ukasp.
Javascript
<script>
// Javascript program for the above approach
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
function sum(arr, N, K)
{
// Stores the sum of the array
var sum = 0;
var v = [];
// Traverse the array arr[]
for (var i = 0; i < N; i++) {
v.push(arr[i]);
}
// Traverse the vector
for (var i = 0;
i < v.length; i++) {
// If v[i] is divisible by K
if (v[i] % K == 0) {
var x = v[i] / K;
// Iterate over the range
// [0, K]
for (var j = 0; j < K; j++) {
// Update v
v.push(x);
}
}
// Otherwise
else
break;
}
// Traverse the vector v
for (var i = 0; i < v.length; i++)
sum = sum + v[i];
// Return the sum of the updated array
return sum;
}
// Driver Code
var arr = [4, 6, 8, 2];
var K = 2;
var N = arr.length;
document.write( sum(arr, N, K));
</script>
Producción:
44
Complejidad temporal: O(N * K * log N)
Espacio auxiliar: O(M), M es el elemento máximo del arreglo .
Enfoque eficiente: el enfoque anterior también se puede optimizar en función de las siguientes observaciones:
- Si arr[i] es divisible por K , entonces sumando arr[i] / K , K veces aumenta la suma en arr[i] .
- Por lo tanto, la idea es empujar el arr[i] / K solo una vez al final del vector.
Siga los pasos a continuación para resolver el problema:
- Inicialice una variable, digamos suma como 0 que almacena la suma de todos los elementos de la array array A[] .
- Inicialice una array, diga A[] y almacene todos los elementos de la array arr[] en A[] .
- Inicialice una variable, digamos marcar como 0 que almacena si el elemento se agregará al final de la array o no.
- Atraviese la array A[] y realice los siguientes pasos:
- Si el indicador de valor es 0 y A[i] es divisible por K , presione A[i] al final de V .
- De lo contrario, actualice el valor de la bandera como 1 .
- Incremente el valor de la suma en V[i % N] .
- Después de completar los pasos anteriores, imprima el valor de la suma como la suma resultante.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
vector<int> v;
// Traverse the array
for (int i = 0; i < N; i++) {
v.push_back(arr[i]);
}
// Stores if the operation
// should be formed or not
bool flag = 0;
// Traverse the vector V
for (int i = 0; i < v.size(); i++) {
// If flag is false and if
// v[i] is divisible by K
if (!flag && v[i] % K == 0)
v.push_back(v[i] / K);
// Otherwise, set flag as true
else {
flag = 1;
}
// Increment the sum by v[i % N]
sum = sum + v[i % N];
}
// Return the resultant sum
return sum;
}
// Driver Code
int main()
{
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout << sum(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
ArrayList<Integer> v = new ArrayList<Integer>();
// Traverse the array
for (int i = 0; i < N; i++) {
v.add(arr[i]);
}
// Stores if the operation
// should be formed or not
boolean flag = false;
// Traverse the vector V
for (int i = 0; i < v.size(); i++) {
// If flag is false and if
// v[i] is divisible by K
if (!flag && v.get(i) % K == 0)
v.add(v.get(i) / K);
// Otherwise, set flag as true
else {
flag = true;
}
// Increment the sum by v[i % N]
sum = sum + v.get(i % N);
}
// Return the resultant sum
return sum;
}
// Driver Code
public static void main (String[] args) {
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = arr.length;
System.out.println(sum(arr, N, K));
}
}
// This code is contributed by Dharanendra L V.
Python3
# Python program for the above approach # Function to calculate sum of array # elements after adding arr[i] / K # to the end of the array if arr[i] # is divisible by K def Sum(arr, N, K): # Stores the sum of the array sum = 0 # Stores the array elements v = [] # Traverse the array for i in range(N): v.append(arr[i]) # Stores if the operation # should be formed or not flag = False i = 0 lenn = len(v) # Traverse the vector V while(i < lenn): # If flag is false and if # v[i] is divisible by K if( flag == False and (v[i] % K == 0)): v.append(v[i]//K) # Otherwise, set flag as true else: flag = True # Increment the sum by v[i % N] sum += v[i % N] i += 1 lenn = len(v) # Return the resultant sum return sum # Driver Code arr = [ 4, 6, 8, 2] K = 2 N = len(arr) print(Sum(arr, N, K)) # This code is contributed by avanitrachhadiya2155
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int []arr, int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
List<int> v = new List<int>();
// Traverse the array
for(int i = 0; i < N; i++)
{
v.Add(arr[i]);
}
// Stores if the operation
// should be formed or not
bool flag = false;
// Traverse the vector V
for(int i = 0; i < v.Count; i++)
{
// If flag is false and if
// v[i] is divisible by K
if (!flag && v[i] % K == 0)
v.Add(v[i] / K);
// Otherwise, set flag as true
else
{
flag = true;
}
// Increment the sum by v[i % N]
sum = sum + v[i % N];
}
// Return the resultant sum
return sum;
}
// Driver Code
static void Main()
{
int[] arr = { 4, 6, 8, 2 };
int K = 2;
int N = arr.Length;
Console.WriteLine(sum(arr, N, K));
}
}
// This code is contributed by SoumikMondal
Javascript
<script>
// javascript program for the above approach
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
function sum(arr, N, K)
{
// Stores the sum of the array
var sum = 0;
var i;
// Stores the array elements
var v = [];
// Traverse the array
for (i = 0; i < N; i++) {
v.push(arr[i]);
}
// Stores if the operation
// should be formed or not
var flag = 0;
// Traverse the vector V
for (i = 0; i < v.length; i++) {
// If flag is false and if
// v[i] is divisible by K
if (!flag && v[i] % K == 0)
v.push(v[i] / K);
// Otherwise, set flag as true
else {
flag = 1;
}
// Increment the sum by v[i % N]
sum = sum + v[i % N];
}
// Return the resultant sum
return sum;
}
// Driver Code
var arr = [4, 6, 8, 2];
var K = 2;
var N = arr.length;
document.write(sum(arr, N, K));
</script>
Producción:
44
Complejidad de Tiempo: O(N * log N)
Espacio Auxiliar: O(N * log N)
Publicación traducida automáticamente
Artículo escrito por IshwarGupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA